How to zip two differently sized lists?
I want to zip two list with different length
for example
A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]
and I expect this
[(1, 'A'), (2, 'B'), (3, 'C'), (4, 'A'), (5, 'B'), (6, 'C'), (7, 'A'), (8, 'B'), (9, 'C')]
But the build-in zip
won't repeat to pair with the list with larger size .
Does there exist any build-in way can achieve this?
thanks
here is my code
idx = 0
zip_list = []
for value in larger:
zip_list.append((value,smaller[idx]))
idx += 1
if idx == len(smaller):
idx = 0
You can use itertools.cycle
:
Make an iterator returning elements from the iterable and saving a copy of each. When the iterable is exhausted, return elements from the saved copy. Repeats indefinitely.
Example:
A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]
from itertools import cycle
zip_list = zip(A, cycle(B)) if len(A) > len(B) else zip(cycle(A), B)
Try this.
A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]
Z = []
for i, a in enumerate(A):
Z.append((a, B[i % len(B)]))
Just make sure that the larger list is in A
.
Do you know that the second list is shorter?
import itertools
list(zip(my_list, itertools.cycle(another_list)))
This will actually give you a list of tuples rather than a list of lists. I hope that's okay.
Solution for an arbitrary number of iterables, and you don't know which one is longest (also allowing a default for any empty iterables):
from itertools import cycle, zip_longest
def zip_cycle(*iterables, empty_default=None):
cycles = [cycle(i) for i in iterables]
for _ in zip_longest(*iterables):
yield tuple(next(i, empty_default) for i in cycles)
for i in zip_cycle(range(2), range(5), ['a', 'b', 'c'], []):
print(i)
Outputs:
(0, 0, 'a', None)
(1, 1, 'b', None)
(0, 2, 'c', None)
(1, 3, 'a', None)
(0, 4, 'b', None)
You can use itertools.cycle
:
from itertools import cycle
my_list = [1, 2, 3, 5, 5, 9]
another_list = ['Yes', 'No']
cyc = cycle(another_list)
print([[i, next(cyc)] for i in my_list])
# [[1, 'Yes'], [2, 'No'], [3, 'Yes'], [5, 'No'], [5, 'Yes'], [9, 'No']]